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A perfectly reflecting mirror has an area of 1 \mathrm{~cm}^2. Light energy is allowed to fall on it for 1 \mathrm{~h} at the rate of 10 \mathrm{W} \mathrm{cm}^{-2}. The force that acts on the mirror is:
 

Option: 1

3.35 \times 10^{-8} \mathrm{~N}




 


Option: 2

6.7 \times 10^{-8} \mathrm{~N}


Option: 3

1.34 \times 10^{-7} \mathrm{~N}


Option: 4

2.4 \times 10^{-4} \mathrm{~N}


Answers (1)

Let, \mathrm{E}= energy falling on the surface per second =10 \mathrm{~J}

Momentum of photons, \mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{\mathrm{h}}{(\mathrm{c} / \mathrm{v})}=\frac{\mathrm{hv}}{\mathrm{c}}=\frac{\mathrm{E}}{\mathrm{c}}

On reflection, change in momentum per second = force

\mathrm{ =2 \mathrm{p}=\frac{2 \mathrm{E}}{\mathrm{c}}=\frac{2 \times 10}{3 \times 10^8}=6.7 \times 10^{-8} \mathrm{~N} }

Hence option 2 is correct.

Posted by

Kshitij

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